Newton’s law of cooling

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By Aditya Abeysinghe

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LAW: The rate of change of the temperature of an object is proportional to the difference between its own temperature and the temperature of its surroundings.

Therefore,dθ / dt = E A (θ – θr ) ; E- A constant that depends upon the object , A – surface area, θ – A certain temperature, θr – Room/ ambient temperature or the temperature of the surroundings.

Newton’s law of cooling

Let’s take the apparatus for this experiment as follows:

Finding the specific heat capacity of a liquid by the method of cooling

Outer vessel

Inner vessel

Cold water

Lid

Strings

Stirer

Thermometer

Calorimeter

Let’s take the ,

Mass of the calorimeter = MMass of water = mw

Mass of the liquid = ml

The specific heat capacity of the calorimeter = CThe specific heat capacity of water = Cw

The specific heat capacity of the liquid = Cl

Also let’s assume that the times taken by water and the liquid to cool from θ1 to θ2 is tw and tl.

Thus, the median rate at which the calorimeter with water looses heat = (MC + mw Cw) (θ1 - θ2 ) / tw

And the median rate at which the calorimeter with water looses heat = (MC + ml Cl) (θ1 - θ2 ) / tl

Under identical situations we can assume that the median rates at which heat is lost by these two systems are equal.

Hence, (MC + mw Cw) (θ1 - θ2 ) / tw = (MC + ml Cl) (θ1 - θ2 ) / tl

If we conduct our experiment, by this equation, we can find the specific heat capacity of the liquid.

1. Set up the apparatus as shown. Fill the empty space between the outer vessel and the inner vessel with cold water.

2. Measure the mass of the calorimeter and pour water of at most 70°C to about 1cm below the lid.

3. Then hang the calorimeter with the two strings, so that it’s free in the air, without any contact with a physical surface.

4. Stir the calorimeter until the temperature of water comes down to about 40°C. Record the temperature at regular intervals.

5. Measure the mass of the calorimeter with water.

The experiment

6. Remove the cold water between the two vessels and refill it with cold water.

7. Remove water inside the calorimeter and repeat the experiment with a similar volume of the liquid.

8. Record the temperatures at regular intervals.

9. Draw the cooling curves for both the liquid and the water in the same graph and within the same temperature range.

θ

t

θ1

θ2

t1 t2

WaterLiquid Use these readings and from the equation we just got above, we can find the specific heat capacity of the liquid

1. The calorimeter should be hanged between the two vessels because during the experiment the environmental conditions can change. Hence the flow of convection currents around the calorimter might change.

2. When similar volumes of the liquid and water are used for the experiment, the temperature variation across the surface of the calorimeter at a certain temperature is identical. Sometimes, when the volumes are different, although inside of the calorimeter might have the same temperture, the temperature distribution across the outer surface might be different.

3. When the calorimeter is stirred, the temperature is equally distributed to all parts of the calorimeter. Thus, the temperature absorbed by convection currents is the same regardless of the place exposed.

Important points

4. When the calorimeter is closed at the top by a lid, the heat loss due to convection and vaporization is minimized.

5. For both occasions, i.e. for the experiment of water and the liquid, the same calorimeter should be used. If we use two identical calorimeters, sometimes the nature of the surface may change. Thus, the readings of the practical and hence the expected value for the specific heat of the gas may change.

7. It’s also better if the calorimeter and the stirer are made of the same material.